## When each of x, y, and z minus the product of the three gives a square

Find rational numbers   $x, \; y, \; z$   such that :

$x \; - \; x \cdot y \cdot z \; = \; A^2$
$y \; - \; x \cdot y \cdot z \; = \; B^2$
$z \; - \; x \cdot y \cdot z \; = \; C^2$

Here’s the first solution:

(1/9),   (441/625),   (1296/1369)   are the first three squares that work.

$x = (1/3)^2 = (1/9)$ ….. $y = (21/25)^2 = (441/625)$ ….. $z = (36/37)^2 = (1296/1369)$

$x \cdot y \cdot z \; = \; (1/9) \cdot (441/625) \cdot (1296/1369) \; = \; 63504/855625 \; = \; (252/925)^2$

$(1/9) \; - \; (252/925)^2 \; = \; 284089/7700625 \; = \; (533/2775)^2$

$(441/625) \; - \; (252/925)^2 \; = \; 21609/34225 \; = \; (147/185)^2$

$(1296/1369) \; - \; (252/925)^2 \; = \; 746496/855625 \; = \; (864/925)^2$

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